Question

Suppose y = f(x) is a differentiable function of x on an interval I and y is one – one, onto and `("d"y)/("d"x)` ≠ 0 on I. Also if f^–1(y) is differentiable on f(I), then `("d"x)/("d"y) = 1/(("d"y)/("d"x)), ("d"y)/("d"x)` ≠ 0

Answer 1

‘y’ is a differentiable function of ‘x’.

Let there be a small increment δx in the value of ‘x’.

Correspondingly, there should be a small increment δy in the value of ‘y’.

As δx → 0, δy → 0

Consider, `(deltax)/(deltay) xx (deltay)/(deltax)` = 1

∴ `(deltax)/(deltay) = 1/((deltay)/(deltax)), (deltay)/(deltax)` ≠ 0

Taking `lim_(deltax -> 0)` on both sides, we get

`lim_(deltax -> 0)((deltax)/(deltay)) = 1/(lim_(deltax -> 0)((deltay)/(deltax))`

Since ‘y’ is a differentiable function of ‘x’,

`lim_(deltax -> 0) ((deltay)/(deltax)) = ("d"y)/("d"x)` and `("d"y)/("d"x)` ≠ 0

∴ `lim_(deltax -> 0)((deltax)/(deltay)) = 1/(("d"y)/("d"x))`

As δx → 0, δy → 0

`lim_(deltax -> 0) ((deltay)/(deltax)) = 1/(("d"y)/("d"x))`  .......(i)

Here, R.H.S. of (i) exist and are finite.

Hence, limits on L.H.S. of (i) also should exist and be finite.

∴ `lim_(deltax -> 0)((deltax)/(deltay)) = ("d"y)/("d"x)`  exists and is finite.

∴ `("d"x)/("d"y) = 1/((("d"y)/("d"x))), ("d"y)/("d"x)` ≠ 0

Alternate Proof:

We know that f^–1[f(x)] = x   .......[Identity function]

Taking derivative on both the sides, we get

`"d"/("d"x) ["f"^-1["f"(x)]] = "d"/("d"x)(x)`

∴ `("f"^-1)"'"["f"(x)]"d"/("d"x)["f"(x)]` = 1

∴ (f^–1)′[f(x)] f′(x) = 1

∴ (f^–1)′[f(x)] = `1/("f""'"(x))`   .......(i)

So, if y = f(x) is a differentiable function of x and x = f^–1(y) exists and is differentiable then

(f^–1)′[f(x)] = (f^–1)′(y) = `("d"x)/("d"y)` and f'(x) = `("d"y)/("d"x)`

∴ Equation (i) becomes

`("d"x)/("d"y) = 1/(("d"y)/("d"x))` where `("d"y)/("d"x)` ≠